## finding three elements in an array whose sum is equal to a given number

Finding three elements in an array whose sum is equal to a given number(say trisum)

Here, for every element(arr[i]) that we pick, trisum-arr[i] gives us the bisum (algorithm to find a pair of elements that matches to sum k), hence we make use of our bisum algorithm to find a pair that matches to sum k

Assumption: The input array is sorted.

Example
arr: {1,2,3,4,5} and trisum is 8 then tri pairs are 1,3,4 and 1,2,5
Here lets say arr[i] is 1 then bisum=trisum-arr[i] i.e, bisum=8-1=7 therefore we should run bisum algorithm for sum 7

Time complexity for this algo: O(n^2) because for every element we need to find set of pairs

Note: If the array is not sorted the the complexity would be nlogn + O(n^2)

```/**
* find 3 elements in an array, that sum up to particular number with best
* complexity
*
* @author ntallapa
*
*/
public class TriPairSumCloseToK {

public void getTriPair(int[] arr, int expectedTriSum) {
for (int i = 0; i < arr.length; i++) {
int expectedBiSum = expectedTriSum - arr[i];
String output = getPair2(arr, expectedBiSum, i);
if(!output.equalsIgnoreCase("")) {
System.out.println(arr[i]+","+output.toString());
}
}
}

/**
* Find the pair in an array whose sum with complexity O(n). This assumes
* the array passed is sorted array and there are no duplicates in the
* array
* @param arr
*            input array of elements
* @param expectedBiSum
*            sum for which pair of array elements needs to be searched
*/
public String getPair2(int[] arr, int expectedBiSum, int ignoreIndex) {
int start = 0;
int end = arr.length - 1;
int sum = 0;

// output array to record matching pairs
StringBuilder arrs = new StringBuilder();

while (start < end) {
//
if(start == ignoreIndex)
start++;
//
if(end == ignoreIndex)
end--;

sum = arr[start] + arr[end];
if (sum == expectedBiSum) {
// we have found one pair, add it to our output array
arrs.append(arr[start] + "," + arr[end] + ";");
start++;
end--;
} else if (sum < expectedBiSum) {
// if the sum of the pair is less than the sum we are searching
// then increment the start pointer which would give us the sum
// more than our current sum as the array is sorted
start++;
} else {
// if the sum of the pair is greater than the sum we are
// searching then decrement the end pointer which would give us
// the sum less than our current sum as the array is sorted
end--;
}
}
return arrs.toString();
}

/**
* Client to test
*
* @param args
*/
public static void main(String[] args) {
TriPairSumCloseToK p = new TriPairSumCloseToK();

// sorted array algo
int[] arr1 = { 1, 2, 3, 4, 5 };
p.getTriPair(arr1, 1);

}
}
```

## algorithm to find first unrepeated character in a string

We can do it in two different ways

• Algorithm with O(2n) time complexity and O(1) space complexity
• Algorithm with O(2n) time complexity and O(n) space complexity

Algorithm with O(2n) time complexity and O(1) space complexity
Here we take a fixed size (256 in which all characters small and caps fall into) integer array and we scan the string two times

• First scan we increment the counter in integer array for every character that appears in string
• In second scan, we look for counter value 1 which gives the first unrepeated character in string

Note: Here space complexity is O(1) because the int array size is always constant irrespective of the string we pass

```	public static char firtUnrepeatedCharInString(char[] s) {
char result = '\0';
int[] counts = new int[256];
for (char c : s) {
counts[c]++;
}

for (int i = 0; i < s.length; i++) {
if (counts[s[i]] == 1) {
result = s[i];
break;
}
}

return result;
}
```

Algorithm with O(2n) time complexity and O(n) space complexity
Here we take hashmap and we scan the string two times

• First scan we insert every character that appears in string as key and its number of appearances as value in hashmap
• In second, scan we look for key whose value is 1 which gives the first unrepeated character in string
```	public static char firtUnrepeatedCharInString(char[] s) {
char result = '\0';
HashMap<Character, Integer> hm = new HashMap<Character, Integer>();
for (char c : s) {
Object o = hm.get(c);
if(o == null) {
hm.put(c, 1);
} else {
int ctr = hm.get(c);
hm.put(c, ++ctr);
}
}

for (char c : s) {
int ctr = hm.get(c);
if(ctr == 1) {
result = c;
break;
}
}
return result;
}
```
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